Question

Dimethyl ether, a useful organic solvent, is prepared in two steps. In the first step, carbon dioxide and hydrogen react to form methanol and water: CO₂(g) + 3H₂(g) $\rightarrow$ CH₃OH (l) + H₂O(l); ΔH = -131.kJ In the second step, methanol reacts to form dimethyl ether and water: 2CH₃OH (l) $\rightarrow$ CH₃OCH₃(g) + H₂O (l); ΔH = 8.kJCalculate the net change in enthalpy for the formation of one mole of dimethyl ether from carbon dioxide and hydrogen from these reactions.Round your answer to the nearest kj.

Answer 1

Answer:

ΔH = - 254 kJ

Explanation:

The final equation for the enthalpy of formation we want to calculate is:

2 CO₂ (g) + 6 H₂(g)   -------------- CH₃OCH₃ (l)     ΔH = ?

Notice that if we take  the 2 times the first reaction and add the second ,  by Hess's law we will add the enthalpy changes to obtain our desired  ΔH .

( CO₂(g) + 3H₂(g)  CH₃OH (l) + H₂O(l); ΔH = -131.kJ)  x 2

2CO₂(g) + 6H₂(g) ---------  2CH₃OH (l) + 2  H₂O(l); ΔH = -262.kJ

2CH₃OH (l)            --------- CH₃OCH₃(g) + H₂O (l); ΔH = 8.kJ

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2CO₂(g) + 6H₂(g)   ----------- CH₃OCH₃(g) + 3H₂O (l); ΔH =  -254 kJ